Atoms, Molecules, and Semiconductor Devices - Nanophysics of Solar and Renewable Energy - page 70

1
= 2 d 2 E

2

d D 2 EðD 2 : 4 Þ¼Eð 2 : 4 Þþ d E = d DðD 2 : 4 Þþ = ðD 2 : 4 Þ þ ... :

ð 3 : 38a Þ

Since the slope is zero at theminimum, that minimumis locally parabolic, which is
the basis for simple harmonic motion, and one sees that the spring constant is
K spring ¼d 2 E /d D 2

¼E 00 . The oscillator frequency then is v¼ ( E 00 / m red ) 1/2 . The value
of E 00 (2.4) ¼ 0.1257 E o / a 2 [40] that gives v¼ ( E 00 / m red ) 1/2 ¼ (0.1257 13.6 1.6 10 19 / 1 / 2

1.67 10 27 ) 1/2 /0.0529 nm ¼ 3.42 10 14 . This corresponds to a zero point energy
h v /2 ¼ 0.113 eV.

3.5
Tetrahedral Bonding in Silicon and Related Semiconductors

The most important linear combinations of angular momentum wavefunctions
(hybrids) cases are sp 3 ,sp 2 , and sp hybrids. sp 3 hybrids describe tetrahedral bonds at
109.5 angles inmethane, CH 4 , and diamond. This scheme also describes bonding in
the important semiconductors Si and GaAs.
In these materials, four outer electrons (two each from 3s and 3p orbitals in Si and
GaAs, and two each from4s and 4p inGe) are stabilized into four tetrahedral covalent
bond orbitals that point from each atom to its four nearest-neighbor atoms, located at
apices of a tetrahedron.
The hybridization effect is essential to understanding molecules and solids. To get
a better understanding, recall (see Table 3.1) that wavefunctions

Y ¼ C 2 r sin

21, 1

e r /2 exp( i j ) are the first two states having angular momentum, where r¼ r / a o

and 21, 1 has a node along z , and

resembles a donut flat in the x -, y -plane. But these wavefunctions can be combined to
make equivalent wavefunctions that point in particular directions.
The sum and difference of these states are also solutions to Schrodingers
equation, and examples are

j is the angle in the x -, y -plane. A polar plot of Y

Y 211 þY 21 1 ¼ C 2 r sin y e r= 2 ½ exp ðijÞþ exp ðijÞ ¼ C 2 r sin y e r= 2 2 cos j ¼ 22 p x ;

ð 3 : 43 Þ

e r= 2 e r= 2 2 sin

Y 211 Y 21 1 ¼ C 2 r sin y ½ exp ði jÞ exp ði jÞ ¼ C 2 ðiÞr sin y j ¼ 22 p y ;

ð 3 : 44 Þ

and

e r= 2 Y 210 ¼ C 2 r cos y ¼ 2 p z : ð 3 : 45 Þ

These linear combinations, 2 p x ,2 p y , and 2 p z , are equivalent to the original
solutions that are eigenstates of angular momentum. They are also a bit like unit
vectors i , j , and k that point along the x -, y -, and z -directions (see Table 3.1 and

Figure 3.1).